Chapter 7.3, Problem 52E

To determine

To evaluate: given expression under the given conditions.

Answer to Problem 52E

  $\sin \frac{\theta }{2}=\frac{1}{\sqrt{26}}$

Explanation of Solution

Given information:

Given expression

  $\sin \left(\theta /2\right);\tan \theta =-\frac{5}{12},$

  $\theta$ in Quadrant IV.

Calculation:

First we Sketch the angle $\theta$ in standard position with terminal side in quadrant III as in figure shown below.

Since $\tan \theta =\frac{-5}{12}$ , we can label a side and the hypotenuse of the triangle in the figure.

  

To find the remaining side, use the Pythagorean theorem. $x^2+y^2=r^2$

On substituting $x=12,y=-5$ , we get

  ${\left(12\right)}^2+{\left(-5\right)}^2=r^2$

Solving for $r$ , we get

  $\begin{array}{c}r=\pm \sqrt{144+25}\\ =\pm 13\end{array}$

Now, $\cos \theta =\frac{x}{r}$ , therefore, we get

  $\cos \theta =\frac{12}{13}$ , note that we have taken positive square root of $r$ because $\cos \theta$ is positive in quadrant IV.

Now, we use the half angle formula for sine,

  $\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\frac{12}{13}}{2}}$

Thus, we have

  $\sin \frac{\theta }{2}=\pm \frac{1}{\sqrt{26}}$

Since $\theta$ lies in quadrant IV, $\frac{\theta }{2}$ lies in quadrant II.

Therefore, we take positive sign of $\sin \frac{\theta }{2}$ .

Hence,

  $\sin \frac{\theta }{2}=\frac{1}{\sqrt{26}}$